JEE Mathematics 2026 Question with Solution

Given:

• $\vec{PQ}=-2\hat i-\hat j+2\hat k$
• $\vec{PR}=a\hat i+b\hat j-4\hat k$, where $a,b\in\mathbb{Z}$
• $|\vec{PR}|=9$
• $\vec{PS}=\hat i-7\hat j+2\hat k$

Find: $3a-4b$

A point equidistant from the two intersecting lines $PQ$ and $PR$ lies on the angle bisector of the angle between them. So $\vec{PS}$ satisfies the angle-bisector condition.

From $|\vec{PR}|=9$,

$$\sqrt{a^2+b^2+16}=9$$

Therefore,

$$a^2+b^2=65$$

Now,

$$\frac{\vec{PS}\cdot\vec{PQ}}{|\vec{PQ}|}=\frac{\vec{PS}\cdot\vec{PR}}{|\vec{PR}|}$$

Compute the left side first:

$$\vec{PS}\cdot\vec{PQ}=(1)(-2)+(-7)(-1)+(2)(2)=-2+7+4=9$$

Also,

$$|\vec{PQ}|=\sqrt{(-2)^2+(-1)^2+2^2}=\sqrt{9}=3$$

So,

$$\frac{\vec{PS}\cdot\vec{PQ}}{|\vec{PQ}|}=\frac{9}{3}=3$$

Hence,

$$\frac{\vec{PS}\cdot\vec{PR}}{9}=3$$

which gives

$$\vec{PS}\cdot\vec{PR}=27$$

Now,

$$\vec{PS}\cdot\vec{PR}=(1)(a)+(-7)(b)+(2)(-4)=a-7b-8$$

So,

$$a-7b-8=27$$

that is,

$$a-7b=35$$

Solving together with $a^2+b^2=65$, we obtain

$$a=7,\quad b=-4$$

Therefore,

$$3a-4b=3(7)-4(-4)=21+16=37$$

So the value of $3a-4b$ is $37$.

The solution concludes with $5$, but its intermediate values are inconsistent with the dot-product condition. Using the shown working correctly gives $37$.

From

$$a^2+b^2=65$$

the integer possibilities are

$$(a,b)=(\pm 1,\pm 8),\ (\pm 8,\pm 1),\ (\pm 4,\pm 7),\ (\pm 7,\pm 4)$$

Also from the angle-bisector condition we got

$$a-7b=35$$

Now test the pair $a=7,\ b=-4$:

$$a-7b=7-7(-4)=7+28=35$$

and

$$a^2+b^2=49+16=65$$

So this pair satisfies both conditions.

Thus,

$$3a-4b=21+16=37$$

Therefore the correct numerical value is $37$.